Problem: What is the area of the region bound by the graphs of $f(x)=x^2+3$, $g(x)=2x+6$, and $x=0$ in quadrant $\text{I}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{11}{3}$ (Choice B) B $9$ (Choice C) C $18$ (Choice D) D $\dfrac{32}{3}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{0}^{b}\left( g(x)-f(x) \right)\,dx$ where $b$ is the $x$ -coordinate of the point of intersection that falls in quadrant $\text{I}$. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ x^2+3&=2x+6\\\\ x^x-2x-3 &= 0 \\\\ (x-3)(x+1) &=0 \end{aligned}$ The graphs intersect where $x=-1$ and $x=3$. Because we only want the part of the graph in quadrant $\text{I}$, we will use $x=3$ as our upper bound. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{0}^{3}\left(2x+6-(x^2+3)\right)\,dx \\\\ &= \int_{0}^{3}\left(-x^2+2x+3\right)\,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{3}\left(-x^2+2x+3\right)\,dx \\\\ &= -\dfrac{x^3}{3}+x^2+3x~\Bigg|_{0}^{3} \\\\ &= \left( -9+9+9 \right) - \left( 0+0+0 \right) \\\\ &= 9 \end{aligned}$ Answer The area is $9$ square units.